Question

solving linear equations (ax + b = c)
in each equation, solve for the variable.
reduce, if necessary, and convert improper fractions to mixed numbers.

1. \\(5x + 3 = 18\\)

\\(x = \underline{\quad}\\)

1. \\(6y - 3 = 0\\)

\\(y = \underline{\quad}\\)

1. \\(7b + 12 = 33\\)

\\(b = \underline{\quad}\\)

1. \\(10n - 7 = 70\\)

\\(n = \underline{\quad}\\)

1. \\(4a + 5 = 41\\)

\\(a = \underline{\quad}\\)

1. \\(9x - 2 = 6\\)

\\(x = \underline{\quad}\\)

1. \\(11y + 6 = 29\\)

\\(y = \underline{\quad}\\)

1. \\(8x - 1 = 4\\)

\\(x = \underline{\quad}\\)

1. \\(7b - 13 = 43\\)

\\(b = \underline{\quad}\\)
challenge: in each equation, solve for the variable. (use mental math if possible.)

1. \\(3a + 7 = -3\\)

\\(a = \underline{\quad}\\)

1. \\(4y + 3 = 0\\)

\\(y = \underline{\quad}\\)

Explanation:

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<reasoning>

Solve equations 1 to 3

1. \(5x + 3 = 18 \implies 5x = 15 \implies x = 3\)
2. \(6y - 3 = 0 \implies 6y = 3 \implies y = \frac{3}{6} = \frac{1}{2}\)
3. \(3b + 12 = 33 \implies 3b = 21 \implies b = 7\)

Solve equations 4 to 6

1. \(10n - 7 = 70 \implies 10n = 77 \implies n = \frac{77}{10} = 7\frac{7}{10}\)
2. \(4a + 5 = 41 \implies 4a = 36 \implies a = 9\)
3. \(9x - 2 = 6 \implies 9x = 8 \implies x = \frac{8}{9}\)

Solve equations 7 to 11

1. \(11y + 6 = 29 \implies 11y = 23 \implies y = \frac{23}{11} = 2\frac{1}{11}\)
2. \(8x - 1 = 4 \implies 8x = 5 \implies x = \frac{5}{8}\)
3. \(7b - 13 = 43 \implies 7b = 56 \implies b = 8\)
4. \(3a + 7 = -3 \implies 3a = -10 \implies a = -\frac{10}{3} = -3\frac{1}{3}\)
5. \(4y + 3 = 0 \implies 4y = -3 \implies y = -\frac{3}{4}\)

</reasoning>

<answer>

No.	Problem	Answer
2	\(6y - 3 = 0\)	\(y = \frac{1}{2}\)
3	\(3b + 12 = 33\)	\(b = 7\)
4	\(10n - 7 = 70\)	\(n = 7\frac{7}{10}\)
5	\(4a + 5 = 41\)	\(a = 9\)
6	\(9x - 2 = 6\)	\(x = \frac{8}{9}\)
7	\(11y + 6 = 29\)	\(y = 2\frac{1}{11}\)
8	\(8x - 1 = 4\)	\(x = \frac{5}{8}\)
9	\(7b - 13 = 43\)	\(b = 8\)
10	\(3a + 7 = -3\)	\(a = -3\frac{1}{3}\)
11	\(4y + 3 = 0\)	\(y = -\frac{3}{4}\)

</answer>

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Answer:

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<reasoning>

Solve equations 1 to 3

1. \(5x + 3 = 18 \implies 5x = 15 \implies x = 3\)
2. \(6y - 3 = 0 \implies 6y = 3 \implies y = \frac{3}{6} = \frac{1}{2}\)
3. \(3b + 12 = 33 \implies 3b = 21 \implies b = 7\)

Solve equations 4 to 6

1. \(10n - 7 = 70 \implies 10n = 77 \implies n = \frac{77}{10} = 7\frac{7}{10}\)
2. \(4a + 5 = 41 \implies 4a = 36 \implies a = 9\)
3. \(9x - 2 = 6 \implies 9x = 8 \implies x = \frac{8}{9}\)

Solve equations 7 to 11

1. \(11y + 6 = 29 \implies 11y = 23 \implies y = \frac{23}{11} = 2\frac{1}{11}\)
2. \(8x - 1 = 4 \implies 8x = 5 \implies x = \frac{5}{8}\)
3. \(7b - 13 = 43 \implies 7b = 56 \implies b = 8\)
4. \(3a + 7 = -3 \implies 3a = -10 \implies a = -\frac{10}{3} = -3\frac{1}{3}\)
5. \(4y + 3 = 0 \implies 4y = -3 \implies y = -\frac{3}{4}\)

</reasoning>

<answer>

No.	Problem	Answer
2	\(6y - 3 = 0\)	\(y = \frac{1}{2}\)
3	\(3b + 12 = 33\)	\(b = 7\)
4	\(10n - 7 = 70\)	\(n = 7\frac{7}{10}\)
5	\(4a + 5 = 41\)	\(a = 9\)
6	\(9x - 2 = 6\)	\(x = \frac{8}{9}\)
7	\(11y + 6 = 29\)	\(y = 2\frac{1}{11}\)
8	\(8x - 1 = 4\)	\(x = \frac{5}{8}\)
9	\(7b - 13 = 43\)	\(b = 8\)
10	\(3a + 7 = -3\)	\(a = -3\frac{1}{3}\)
11	\(4y + 3 = 0\)	\(y = -\frac{3}{4}\)

</answer>

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